The n'th prime looks a bit like n log n. (This is basically the same fact as the "prime number theorem": the number of primes up to n is roughly n/log(n).)
We have:
sum (up to n) of 1/x ~= log n [harmonic series]
sum (up to n) of 1/(x log x) ~= log log n [reciprocals of primes, roughly]
sum (up to n) of 1/(x log x log log x) ~= log log log n
and so on. All these series diverge, but slower and slower and slower. And clearly these series "only just" diverge. In fact, iIf you take, e.g.,
sum of 1/(x log x log log x (log log log x)^1.1)
then it converges. If you have just the right kind of warped mind, the question that will immediately occur to you on contemplating this is: what if you define L(x) := x log x log log x log log log x ... continuing the product up to, but excluding, the first factor that's < 1, and ask about the sum of 1/L(x)?
Answer: The series diverges really slowly. The points at which L(x) starts to use one extra logarithm are e, e^e, e^e^e, e^e^e^e, etc. The sums between each adjacent pair of these are comparable in size. So the sum up to n is roughly the number of times you need to take logs, starting with n, before you get down to 1.
[EDITED to add: The first bit of the above is closely related to cperciva's comment that's a sibling of this one.]
We have:
sum (up to n) of 1/x ~= log n [harmonic series]
sum (up to n) of 1/(x log x) ~= log log n [reciprocals of primes, roughly]
sum (up to n) of 1/(x log x log log x) ~= log log log n
and so on. All these series diverge, but slower and slower and slower. And clearly these series "only just" diverge. In fact, iIf you take, e.g.,
sum of 1/(x log x log log x (log log log x)^1.1)
then it converges. If you have just the right kind of warped mind, the question that will immediately occur to you on contemplating this is: what if you define L(x) := x log x log log x log log log x ... continuing the product up to, but excluding, the first factor that's < 1, and ask about the sum of 1/L(x)?
Answer: The series diverges really slowly. The points at which L(x) starts to use one extra logarithm are e, e^e, e^e^e, e^e^e^e, etc. The sums between each adjacent pair of these are comparable in size. So the sum up to n is roughly the number of times you need to take logs, starting with n, before you get down to 1.
[EDITED to add: The first bit of the above is closely related to cperciva's comment that's a sibling of this one.]